Question

How to solve ${{a}^{3}}-3a+2=0$ ?

Answer 1

Hint: We are given ${{a}^{3}}-3a+2=0$, we have to solve this, we will understand what the solution means then we look at the degree as it will tell us about the number of solutions. Then we will find the factor form of ${{a}^{3}}-3a+2=0$, once we achieve that then using product zero rule we will achieve the required solution of the problem. We can also cross check if our solutions are correct or not.

Complete step-by-step solution:
We are given ${{a}^{3}}-3a+2=0$, we start by learning any equation. solution in those values of the variable which when inserted in the given equation then we get that equation is satisfied.
Now the degree of the equation tells us about the possible number of solutions of any equation in ${{a}^{3}}-3a+2=0$, degree is ‘3’. So, this can have a maximum of 3 solutions at most .
Now we will try to split our equation and write it into factored form. Before we start we know adding and subtraction will never affect our equation.
So, adding and subtraction of ${{a}^{2}}$ to ${{a}^{3}}-3a+2=0$we get –
${{a}^{3}}-{{a}^{2}}+{{a}^{2}}-3a+2=0$
Now as $-3a=-a-2a$ so using this above, we get –
${{a}^{3}}-{{a}^{2}}+{{a}^{2}}-a-2a+2=0$
Now we make a pair of 2 terms and take out common terms.
So, $\left( {{a}^{3}}-{{a}^{2}} \right)+\left( {{a}^{2}}-a \right)-\left( 2a-2 \right)=0$
Taking common terms, we get –
${{a}^{2}}\left( a-1 \right)+a\left( a-1 \right)-2\left( a-1 \right)=0$
As $\left( a-1 \right)$ is the common factor of each term , we can write it as –
$\left( a-1 \right)\left( {{a}^{2}}+a-2 \right)=0$
Now we solve ${{a}^{2}}+a-2$ further
We use a middle term split method to split it further.
So, as we have middle term as 1 in ${{a}^{2}}+a-2$
So we can write $1=2-1$ . So,
$\begin{align}
  & {{a}^{2}}+a-2={{a}^{2}}+\left( 2-1 \right)a-2 \\
 & ={{a}^{2}}+2a-a-2 \\
\end{align}$
Taking common term, we get –
$=a\left( a+2 \right)-1\left( a+2 \right)$
As at 2 is common so we get –
$=\left( a-1 \right)\left( a+2 \right)$
So we get –
${{a}^{2}}+a-2=\left( a-1 \right)\left( a+2 \right)$
Hence we get –
${{a}^{3}}-3a+2=0$
Become
$\left( a-1 \right)\left( a-1 \right)\left( a-2 \right)=0$
Now using zero product rules which say if $a\times b\times c=0$ then either ‘a’ or ‘b’ or ‘c’ is zero.
So
$a-1=0\text{ or }a+2=0$
By simplifying, we get –
$a-1\text{ or }a=-2$
So we get –
$a=1$ and $a=-2$ are the solution.
Here $a=1$ is solution of multiplicity 2
And $a=-2$ is solution of multiplicity 1

Note: We can cross check that $a=1\text{ and }-2$ are solution or not
We put $a=1$ and $a=-2$ one by one and check
Now putting $a=1$ in ${{a}^{3}}-3a+2=0$
We get –
$\begin{align}
  & {{1}^{3}}-3\times 1+2=0 \\
 &\Rightarrow 1-3+2=0 \\
 &\Rightarrow 3-3=0 \\
 &\Rightarrow 0=0 \\
\end{align}$
Which is true
So $a=1$ is the solution
This is true.
Now, putting $a=-2$ , in ${{a}^{3}}-3a+2=0$ we get –
$\begin{align}
  & {{\left( -2 \right)}^{3}}-3\left( -2 \right)+2=0 \\
 &\Rightarrow -8+6+2=0 \\
 &\Rightarrow -8+8=0 \\
 &\Rightarrow 0=0 \\
\end{align}$
Which is true
So, $a=-2$ is the solution.
Hence our solution is correct.